A formula for solving differential equations of the form $\frac{du}{dt}= Au$?

Question

Notation

• $\lambda_{1,1}, \lambda_{2,2}, \ldots, \lambda_{n,n}$ are all the eigenvalues of $A_n$.
• $x_1, x_2, \ldots, x_n$ are all the eigen vectors of $A_n$.
• $\Lambda_n = \text{diag}\left(\lambda_{1,1}, \lambda_{2,2}, \ldots, \lambda_{n,n}\right)$.
• $S = \left[\begin{array}{cccc} | & | & & | \\ x_1 & x_2 & \cdots & x_n \\ | & | & & | \end{array}\right]$.

Question

q1: Is the solution to $\frac{du}{dt} = Au$, given $u(0)$, correctly described by...

$$ u(t) = \sum_{k = 1}^{n} \left\{\mathrm{e}^{\lambda_{k,k}t}c_kx_k \right\} \style{font-family:inherit}{\text{ s.t. }} Sc = u(0)\tag{$\style{font-family:inherit}{\text{e1}}$}\label{eq1} $$

Example

Given that...

• $A_5 = \text{ones}\left(5\right)$.
• $u(0) = \left[\begin{array}{ccccc} 0 & 1 & 1 & 1 & 2 \end{array}\right]^T$.

We determine that...

• $\Lambda = \left[ \begin{array}{ccccc} 5 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$.
• $S = \left[ \begin{array}{ccccc} 1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \end{array}\right]$.
• $Sc = u(0) \Longrightarrow c = \left[\begin{array}{ccccc} 1 & 1 & 0 & 0 & 0 \end{array}\right]^T$.

Using $\ref{eq1}$...

$$ u(t) = \sum_{k = 1}^{5} \left\{\mathrm{e}^{\lambda_{k,k}t}c_kx_k \right\}\\ \Longrightarrow u(t) = \mathrm{e}^{(5)t}(1)x_1 + \mathrm{e}^{(0)t}(1)x_2 + \mathrm{e}^{(0)t}(0)x_3 + \mathrm{e}^{(0)t}(0)x_4 + \mathrm{e}^{(0)t}(0)x_5\\ \Longrightarrow u(t) = \mathrm{e}^{5t}\left[\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{array}\right] + \left[\begin{array}{c} -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right] $$

Additional Questions

q2: If your answer to q1 was yes, is there a way to express $\ref{eq1}$ more compactly?
q3: Please re-answer q1 but wrt the following (attempt @ answering q2)...

$$ u(t) = \frac{S\text{e}^{\Lambda t}\text{adj}\left(S\right)u(0)}{|S|}\tag{$\style{font-family:inherit}{\text{e2}}$}\label{eq2} $$

I verified that e1 & e2 yield the same answer to my example problem (but they might not work for others?), using Mathematica...

adj[m_] := 
    Map[Reverse, Minors[Transpose[m], Length[m] - 1], {0, 1}] * 
      Table[(-1)^(i + j), {i, Length[m]}, {j, Length[m]}];

A = ConstantArray[1,{5,5}];
\[CapitalLambda] = DiagonalMatrix[Eigenvalues[A]];
S = Transpose[Eigenvectors[A]];
u0 = {0,1,1,1,2};
c = LinearSolve[S,u0];

(* Answer w/ method 1 *)
a1 = Sum[Exp[Eigenvalues[A][[k]]]*c[[k]]*Eigenvectors[A][[k]],{k,1,Length[S]}];

(* Answer w/ method 2 *)
a2 = (S.MatrixExp[\[CapitalLambda]*t].adj[S].u0)/Det[S];

Print["A = ", A//MatrixForm];
Print["\[CapitalLambda] = ", \[CapitalLambda]//MatrixForm];
Print["S = ", S//MatrixForm];
Print["u(0) = ", u0//MatrixForm];
Print["c = ", c//MatrixForm];
Print["a1: u(t) = ", a1//MatrixForm];
Print["a2: u(t) = ", a2//MatrixForm];

Answer 1

Yes. The exponential of a square matrix $A$ (or, more generally, of a bounded linear operator $A$ on a normed vector space) is defined by the Taylor series: $$ \exp(A) = \sum_{k \geq 0} {1 \over k!} A^{k}. $$ Notice that this is also a square matrix (of the same dimension as $A$). How would we express its eigenvalues in terms of those of $A$?

The solution to the IVP $$ \dot{u} = Au, \quad u(0) = u_0 $$ is: $$ u(t) = \exp(t A) \, u_{0}, $$ which can be verified directly. All that remains is to express $u_0$ as a linear combination of the eigenvectors of $\exp(tA)$, and to figure out how the eigenvalues of $\exp(tA)$ are related to those of $A$.